How do you find the sum of the infinite geometric series given $Sigma(1.5)(0.25)^(n-1)$ from n=1 to $oo$ ?

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Answer 1 · 2016-11-23T05:03:32

$Sigma_(n->oo)(1.5)(0.25)^(n-1)=2$

As in expanded form this series $Sigma(1.5)(0.25)^(n-1)$ is

$1.5+(1.5)(0.25)+(1.5)(0.25)^2+(1.5)(0.25)^3+(1.5)(0.25)^4+ .......$ ,

it is a geometric series whose first term $a_1$ is $1.5$ and common ratio $r$ is $0.25$ .

As $r<1$ , the sum of infinite series is $a/(1-r)=1.5/(1-0.25)=1.5/0.75=2$

Hence $Sigma_(n->oo)(1.5)(0.25)^(n-1)=2$

How do you find the sum of the infinite geometric series given Sigma(1.5)(0.25)^(n-1)Sigma(1.5)(0.25)^(n-1) from n=1 to oooo?